Physics, asked by param7777777, 11 months ago

Q.11 A particle of mass 3 kg moves under the force of
4i+8j+10k N. If the particle starts from rest
and was at origin initially. Its new co-ordinates
after 3 second is-
(1) (4, 8, 10) (2) (6, 12, 15)
(3) (2, 4,5) (4) (3, 6, 7.5)​

Answers

Answered by QGP
111

Kinematics in 3-dimensions

To solve the problem, we can consider the motion independently along all the three dimensions.

Mass m = 3 kg

Force \vec{F} = 4\hat{\imath}+8\hat{\jmath}+10\hat{k}

Time t = 3 s

Initial Velocity u = 0 m/s

Consider the motion individually along the x, y and z directions.

 \rule{320}{1}

The X-Direction

Here, F_x = 4\ N

So, the acceleration would be

a_x = \dfrac{F_x}{m} = \dfrac{4}{3}\ m/s^2

Hence, the displacement along x-direction can be calculated as:

x = u_xt+\dfrac{1}{2}a_xt^2 \\\\\\ \implies x=0+\dfrac{1}{2}\times \dfrac{4}{3}\times 3^2 \\\\\\ \implies \boxed{x = 6\ m}

[Note that we have already got the Answer. Option 2: (6, 12, 15). We will still solve the question for the sake of completion.]

 \rule{320}{1}

The Y-Direction

Here, F_y = 8\ N

So, the acceleration would be

a_y = \dfrac{F_y}{m} = \dfrac{8}{3}\ m/s^2

Hence, the displacement along y-direction can be calculated as:

y = u_yt+\dfrac{1}{2}a_yt^2 \\\\\\ \implies y=0+\dfrac{1}{2}\times \dfrac{8}{3}\times 3^2 \\\\\\ \implies \boxed{y = 12\ m}

 \rule{320}{1}

The Z-Direction

Here, F_z = 10\ N

So, the acceleration would be

a_z = \dfrac{F_z}{m} = \dfrac{10}{3}\ m/s^2

Hence, the displacement along z-direction can be calculated as:

z = u_zt+\dfrac{1}{2}a_zt^2 \\\\\\ \implies z=0+\dfrac{1}{2}\times \dfrac{10}{3}\times 3^2 \\\\\\ \implies \boxed{x = 15\ m}

 \rule{320}{1}

Thus, we have our new co-ordinates:

x = 6 m

y = 12 m

z = 15 m

Hence, The Answer is Option 2: (6, 12, 15)


Anonymous: Ñîçê Âñswêr:)
QGP: Thank You :)
Answered by Anonymous
68

Hey there

refer to attachment

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